In the figure below, a 36.0 kg uniform square sign, of edge L = 2.00 m, is hung
ID: 1443680 • Letter: I
Question
In the figure below, a 36.0 kg uniform square sign, of edge L = 2.00 m, is hung from a horizontal rod of length d_n = 3.00 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance d_v = 4.00 m above the point where the rod is hinged to the wall. (a) What is the tension in the cable? N (b) What are the magnitude and direction of the horizontal component of the force on the rod from the wall? (Include the sign. Take the positive direction to be to the right.) N (c) What are the magnitude and direction of the vertical component of this force? (Include the sign. Take the positive direction to be upward.) NExplanation / Answer
a) tension in the cable is T
vertical component of tension = Tsin(theta)
where tan(theta) = dv/dh = 4/3
=> sin(theta) = 4/5 = 0.8
NET torque is zero about hinge
=> Tsin(theta)* dh = mg*(dh - L/2)
=> T*0.8*3.0 = 36*9.8*(3.0 - 1/2 )= 36*9.8*2.5 = 882
=> T = 367.5 N
b) horizontal component of force on the rod
Fh = Tcos(theta) = 367.5*0.6 = 612.5 N
c) vertical component of force
Fv = Tsin(theta) = 367.5*0.8 = 294 N